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natta225 [31]
3 years ago
6

Find the inequality represented by the graph​

Mathematics
2 answers:
Vika [28.1K]3 years ago
8 0

Answer:

y < 1/4x + 3

Step-by-step explanation:

the line is dotted so we know the symbol will not include the "equal to" part

we can also see that the shading in below the y intercept, so we know that the symbol will also be less than

this means the symbol is <

now, create an equation for the line:

we know that the y intercept is 3, and that there is a rise of 1 and a walk of 4

so the slope is 1/4

now put this together and you get 1/4x + 3

this part will go on the right side of the < symbol in your inequality

so now you can create an inequality: y < 1/4x + 3

Ahat [919]3 years ago
8 0

Answer:

<em>y < </em>\frac{1}{4}<em> x + 3 </em>

Step-by-step explanation:

m = \frac{1}{4} and coordinates of y-intercept are (0, 3)

<em>y < </em>\frac{1}{4}<em> x + 3</em>

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Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

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Duck #1 lays eggs whose weights are normally distributed with a mean of 70gramsand a standard deviation of 6 grams.Duck #2 lays
Genrish500 [490]

Answer:

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If an egg is randomly chosen from each duck, what is the probability that Duck #2’s egg weighs more than Duck #1’s egg?

Duck #2 egg will weigh more if the subtraction of duck's 2 egg by duck's 1 egg is larger than 0.

When we subtract normal distributions, the mean is the subtraction of the means. So

\mu = 65 - 70 = -5

The standard deviation is the square root of the sum of the variances. So

\sigma = \sqrt{6^2+5^2} = \sqrt{61} = 7.81

Now, we have to find 1 subtracted by the pvalue of Z when X = 0. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{0 - (-5)}{7.81}

Z = 0.64

Z = 0.64 has a pvalue of 0.7389

1 - 0.7389 = 0.2611

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

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3 years ago
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